Question

The feet of the average adult woman are 24.6 cm long, and foot lengths are normally distributed. If 16% of adult women have feet that are shorter than 22 cm, approximately what percent of adult women have feet longer than 27.2 cm?

Answer 1

Approximately 16% of adult women have feet longer than 27.2 cm.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The feet of the average adult woman are 24.6 cm long

This means that
`\mu = 24.6`

16% of adult women have feet that are shorter than 22 cm

This means that when
`X = 22`, Z has a p-value of 0.16, so when
`X = 22, Z = -1`. We use this to find
`\sigma`

`Z = (X - \mu)/(\sigma)`

`-1 = (22 - 24.6)/(\sigma)`

`-\sigma = -2.6`

`\sigma = 2.6`

Approximately what percent of adult women have feet longer than 27.2 cm?

The proportion is 1 subtracted by the p-value of Z when X = 27.2. So

`Z = (X - \mu)/(\sigma)`

`Z = (27.2 - 24.6)/(2.6)`

`Z = 1`

`Z = 1` has a p-value of 0.84.

1 - 0.84 = 0.16

0.16*100% = 16%.

Approximately 16% of adult women have feet longer than 27.2 cm.